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MSC 81V70, 35Q55, 46N50
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Bose systems, ground state properties
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\begin{document}
\title{{\bf The Ground State Energy of a Dilute Bose Gas}\thanks{\copyright
1999 by the authors. Reproduction
of this work, in its entirety, by any means, is permitted for
non-commercial purposes.}\thanks
{For the Proceedings of the International Conference on Partial Differential
Equations and Mathematical Physics, Birmingham, Alabama, March 15--19, 1999}}
\author{Elliott H. Lieb\thanks{Supported in part
by NSF Grant PHY-98 20650.}\\
Departments of Mathematics and Physics\\
Princeton University, Princeton, New Jersey 08544-0708\\ {\it
lieb@math.princeton.edu}
\and
Jakob Yngvason\thanks{Supported in part by
the Adalsteinn Kristj\'ansson Foundation, University of Iceland.}\\ Institut
f\"ur Theoretische Physik,
Universit\"at
Wien,\\Boltzmanngasse 5, A 1090 Vienna, Austria\\ {\it
yngvason@thor.thp.univie.ac.at}}
%\date{October 15, 1999}
\maketitle
% General info
%\subjclass{Primary 81V10, 81V70; Secondary 81V45, 81T08, 81T16}
\date{today}
\begin{abstract} According to a formula that was put forward many decades ago
the ground state energy per particle of an interacting, dilute Bose gas at
density $\rho$ is $2\pi\hbar^2\rho a/m$ to leading order in $\rho a^3\ll 1$,
where $a$ is the scattering length of the interaction potential and $m$ the
particle mass. This result, which is important for the theoretical
description
of current experiments on Bose-Einstein condensation, has recently been
established rigorously for the first time. We give here an account of the
proof
that applies to nonnegative, spherically symmetric potentials decreasing
faster
than $1/r^3$ at infinity. \end{abstract}
\maketitle
%%%%% Type your manuscript here %%%%%
\section{Introduction}
Recent progress in the trapping and cooling of atoms has
made the ground state properties of dilute, interacting Bose gases accessible
to
experimental study \cite{TRAP}, \cite{DGPS}. In the theoretical description of
such experiments an old
formula for the ground state energy plays an important role. This formula,
stated precisely in (\ref{basic}) below, is the subject of the present
contribution, which is
essentially an exposition of the paper \cite{LY1998},
incorporating some new results from \cite{LSY1999} and \cite{S1999}.
We consider the Hamiltonian for $N$ Bosons of mass
$m$ enclosed in a cubic box $\Lambda$ of side length $L$ and interacting by a
spherically symmetric pair potential
$v(|\x_i - \x_j|)$:
\begin{equation}\label{ham}
H_{N} = - \mu\sum_{i=1}^{N} \Delta_i +
\sum_{1 \leq i < j \leq N} v(|\x_i - \x_j|).
\end{equation}
Here $\x_i\in\mathbb R^3$, $i=1,\dots,N$ are the positions of the particles,
$\Delta_i$ the Laplacian with respect to $\x_{i}$,
and we have denoted ${\hbar^2}/{ 2m}$ by $\mu$ for short. (By choosing
suitable units $\mu$ could, of course, be eliminated, but we want to keep
track
of the dependence of the energy on Planck's constant and the mass.) The
Hamiltonian
(\ref{ham}) operates on {\it symmetric} wave functions in $L^2(\Lambda^{N},
d\x_1\cdots d\x_N)$ as is appropriate for Bosons. The interaction
potential will be assumed to be {\it nonnegative} and to decrease
faster than $1/r^3$ at infinity.
We are interested in the ground state energy $E_{0}(N,L)$ of (\ref{ham}) in
the
{\it thermodynamic limit} when $N$ and $L$ tend to infinity with the
density $\rho=N/L^3$ fixed. The energy per particle in this limit
%
\begin{equation} e_{0}(\rho)=\lim_{L\to\infty}E_{0}(\rho L^3,L)/(\rho
L^3).\end{equation}
%
Our results about $e_{0}(\rho)$ are based on estimates on
$E_{0}(N,L)$
for finite $N$ and $L$, which are important, e.g., for the considerations of
inhomogeneous systems in \cite{LSY1999}.
To define $E_{0}(N,L)$ precisely one
must specify the boundary conditions. These should not matter for the
thermodynamic limit.
To be on the safe side we use Neumann boundary conditions for the
lower bound, and Dirichlet boundary conditions for the upper bound
since these lead, respectively, to the lowest and the highest energies.
For experiments with dilute gases the {\it low density asymptotics} of
$e_{0}(\rho)$ is of importance. Low density means here that the mean
interparticle distance, $\rho^{-1/3}$ is much larger than the
{\it scattering length} $a$ of the potential, defined as
\begin{equation}a=\lim_{r\to\infty}r-\frac{u_{0}(r)}{u_{0}'(r)},\end{equation}
where $u_{0}$ solves the zero energy scattering equation,
%
\begin{equation}\label{scatteq}
-2\mu u_{0}^{\prime\prime}(r)+ v(r) u_{0}(r)=0
\end{equation}
%
with $u_{0}(0)=0$. (The factor $2$ in (\ref{scatteq}) comes from
the reduced mass of the two particle problem.) Our main result is a
rigorous proof of the formula
\begin{equation} e_{0}(\rho)\approx4\pi\mu\rho a\end{equation}
for $\rho a^3\ll 1$, more precisely of
\begin{theorem}[Low density limit of the ground state energy]
\begin{equation}\label{basic}
\lim_{\rho a^3\to 0}\frac {e_{0}(\rho)}{4\pi\mu\rho a}=1.
\end{equation}
\end{theorem}
This formula is independent of the boundary conditions used for the
definition of $e_{0}(\rho)$.
The genesis of an understanding of $e_{0}(\rho)$ was the pioneering
work \cite{BO} of Bogol\-iubov, and in the 50's and early 60's several
derivations of (\ref{basic}) were presented \cite{Lee-Huang-YangEtc},
\cite{Lieb63}, even including higher order terms:
\begin{equation}\frac{e_{0}(\rho)}{4\pi\mu\rho a}=
1+\mfr{128}/{15\sqrt \pi}(\rho a^3)^{1/2}
+8\left(\mfr{4\pi}/{3}-\sqrt 3\right)(\rho a^3)\log (\rho a^3)
+O(\rho a^3)
\end{equation}
These early developments are reviewed in \cite{EL2}. They all rely
on some special assumptions about the ground state that have never been
proved, or on the selection of special terms from a perturbation series
which likely diverges. The only rigorous estimates of this period were
established by Dyson, who derived the following bounds in 1957 for a
gas of hard spheres \cite{dyson}:
\begin{equation} \frac1{10\sqrt 2} \leq
\frac{e_{0}(\rho)}{ 4\pi\mu\rho a}\leq\frac{1+2 Y^{1/3}}{
(1-Y^{1/3})^2}
\end{equation}
with $Y=4\pi\rho a^3/3$. While the upper bound has the asymptotically
correct form, the lower bound is off the mark by a factor of about 1/14.
But for about 40 years this was the best lower bound available!
Since (\ref{basic}) is a basic result about the Bose gas it is clearly
important to derive it rigorously and in reasonable generality, in
particular for more general cases than hard spheres. The
question immediately arises for which interaction potentials one
may expect it to be true. A notable fact is that it
{\it not true for all} $v$ with $a>0$, since there are two body
potentials with positive scattering length that allow many body bound
states \cite{BA}. Our proof, presented in the sequel, works for nonnegative
$v$, but we
conjecture that (\ref{basic}) holds if $a>0$ and $v$ has no $N$-body bound
states for any $N$. The lower bound is, of course, the hardest part, but the
upper bound is not altogether trivial either.
%%%%%%%%
Before we start with the estimates a simple computation and some
heuristics may be helpful to make
(\ref{basic}) plausible and motivate the formal proofs.
With $u_{0}$ the scattering solution and
$f_{0}(r)=u_{0}(r)/r$,
partial integration gives
\begin{eqnarray}\label{partint}
\int_{|\x|\leq R}\{2\mu|\nabla f_{0}|^2+v|f_{0}|^2\}d\x&=&
4\pi\int_{0}^{R}\{2\mu[u_{0}'(r)-(u_{0}(r)/r)]^2+v(r)|u_{0}(r)]^2\}dr
\nonumber\\&=&
8\pi\mu a |u_{0}(R)|^2/R^2\to 8\pi\mu a\quad\mbox{\rm for $R\to\infty$},
\end{eqnarray}
if $u_{0}$ is normalized so that $f_{0}(R)\to 1$ as $R\to\infty$.
Moreover, for positive interaction potentials the scattering solution
minimizes
the quadratic form in (\ref{partint}) for each $R$ with
$u_{0}(0)=0$ and $u_{0}(R)$ fixed as boundary conditions. Hence the energy
$E_{0}(2,L)$ of two
particles in a large box, i.e., $L\gg
a$, is approximately $8\pi\mu a/L^3$. If the gas is sufficiently
dilute it is not unreasonable to expect that the energy is essentially
a sum of all such two particle contributions. Since there are
$N(N-1)/2$ pairs, we are thus lead to $E_{0}(N,L)\approx 4\pi\mu a
N(N-1)/L^3$, which gives (\ref{basic}) in the thermodynamic limit.
This simple heuristics is far from a rigorous proof, however,
especially for the lower bound. In fact, it is rather remarkable that
the same asymptotic formula holds both for `soft' interaction
potentials, where perturbation theory can be expected to be a good
approximation, and potentials like hard spheres where this is not so.
In the former case the ground state is approximately the constant
function and the energy is {\it mostly potential}:
According to perturbation theory
$E_{0}(N,L)\approx N(N-1)/(2 L^3)\int v(|\x|)d\x$. In particular it is {\it
independent of} $\mu$, i.e. of Planck's constant and mass. Since,
however, $\int v(|\x|)d\x$ is the first Born approximation to $8\pi\mu
a$ (note that $a$ depends on $\mu$!), this is not in conflict with
(\ref{basic}).
For `hard' potentials on the other hand, the ground state is {\it
highly correlated}, i.e., it is far from being a product of single
particle states. The energy is here {\it mostly kinetic}, because the
wave function is very small where the potential is large. These two
quite different regimes, the potential energy dominated one and the
kinetic energy dominated one, cannot be distinguished by the low
density asymptotics of the energy. Whether they behave
differently with respect to other phenomena, e.g., Bose-Einstein
condensation, is not known at present.
Bogolubov's analysis \cite{BO} presupposes the existence of Bose-Einstein
condensation. Nevertheless, it is correct (for the energy) for the
one-dimensional delta-function Bose gas \cite{LL}, despite the fact
that there is (presumably) no condensation in that case. It turns
out that BE condensation is not really needed in order to understand
the energy. As we shall see, `global' condensation can be replaced
by a `local' condensation on boxes whose size is independent of
$L$. It is this crucial understanding that enables us to prove Theorem
1.1 without having to decide about BE condensation.
An important idea of Dyson was to transform the hard sphere
potential into a soft potential at the cost of sacrificing the
kinetic energy, i.e., effectively to move from one
regime to the other. We shall make use of this idea in our proof
of the lower bound below. But first we discuss the simpler upper
bound, which relies on other ideas from Dyson's beautiful paper \cite{dyson}.
%%%%%%%%%%%
\section{Upper bound}
The following generalization of Dyson's upper bound holds
\cite{LSY1999}, \cite{S1999}:
\begin{theorem}[Upper bound] Define $\rho_{1}=(N-1)/L^3$ and
$b=(4\pi\rho_{1}/3)^{-1/3}$. For nonnegative potentials $v$, and $b>a$
the ground state energy of (\ref{ham}) with periodic boundary conditions
satisfies
\begin{equation}\label{upperbound}
E_{0}(N,L)/N\leq 4\pi \mu \rho_{1}a\frac{1-\frac{a}
{b}+\left(\frac{a}
{b}\right)^2+\frac12\left(\frac{a}
{b}\right)^3}{\left(1-\frac{a}
{b}\right)^8}.
\end{equation}
For Dirichlet boundary conditions the estimate holds with ${\rm
(const.)}/L^2$ added to the right side.
Thus in the thermodynamic limit and for all boundary conditions
\begin{equation}
\frac{e_{0}(\rho)}{4\pi\mu\rho a}\leq\frac{1-Y^{1/3}+Y^{2/3}-\mfr1/2Y}
{(1-Y^{1/3})^8}.
\end{equation}
provided $Y=4\pi\rho a^3/3<1$.
\end{theorem}
{\it Remark.} The bound (\ref{upperbound}) holds for potentials
with infinite range, provided $b>a$. For potentials of finite range
$R_{0}$ it can be improved for $b>R_{0}$ to
\begin{equation}\label{upperbound2}
E_{0}(N,L)/N\leq 4\pi \mu \rho_{1}a\frac{1-\left(\frac{a}
{b}\right)^2+\frac12\left(\frac{a}
{b}\right)^3}{\left(1-\frac{a}
{b}\right)^4}.
\end{equation}
{\it Proof.}
We first remark that the expectation value of (\ref{ham}) with any
trial wave function gives an upper bound to the bosonic ground state
energy, even if the trial function is not symmetric under permutations
of the variables. The reason is that an absolute ground state of the
elliptic differential operator (\ref{ham}) (i.e. a ground state
without symmetry requirement) is a nonnegative function which can be
be symmetrized without changing the energy because (\ref{ham}) is
symmetric under permutations. In other words, the absolute ground
state energy is the same as the bosonic ground state energy.
Following \cite{dyson} we choose a trial function of
the following form
\begin{equation}\label{wave}
\Psi(x_1,\dots,x_N)=F_1(x_{1}) \cdot F_2(x_{1},x_{2}) \cdots
F_N(x_{1},\dots,x_{n}).
\end{equation}
More specifically, $F_{1}\equiv 1$ and $F_{i}$
depends only on the distance of $x_{i}$ to its nearest neighbor among
the the points $x_1,\dots ,x_{i-1}$ (taking the periodic boundary
into account):
\begin{equation}\label{form}
F_i(x_1,\dots,x_i)=f(t_i),
\quad t_i=\min\left(\xij,j=1,\dots,
i-1\right),
\end{equation}
with a function $f$ satisfying
\begin{equation}0\leq
f\leq 1, \quad f'\geq 0.
\end{equation}
The intuition behind the ansatz (\ref{wave}) is that the particles are
inserted into the system one at the time, taking into account the
particles previously inserted. While such a wave function cannot
reproduce all correlations present in the true ground state, it turns
out to capture the leading term in the energy for dilute gases.
The form (\ref{form}) is computationally easier to handle than an
ansatz of the type $\prod_{ib$},
\end{cases}
\end{equation}
with $f_{0}(r)=u_{0}(r)/r$. The estimates (\ref{upperbound}) and
(\ref{upperbound2}) are
obtained by somewhat lengthy computations similar as in
\cite{dyson}, but making use of
(\ref{partint}). For details we refer to \cite{LSY1999} and \cite{S1999}.
A test wave function with Dirichlet boundary condition may be obtained
by localizing the wave function (\ref{wave}) on the length scale $L$.
The energy cost per particle for this is ${\rm (const.)}/L^2$.
\hfill$\Box$
\section{Lower bound}
%%%%%%%%%%%%%%%%%%%%%%
To get an idea why the lower bound
for the bosonic ground state energy of (\ref{ham}) is not easy to
obtain let us consider the relevant length scales of the problem.
These are
\begin{itemize}
\item The scattering length $a$.
\item The mean particle distance $\rho^{-1/3}$.
\item The `uncertainty principle length' $\ell_{c}$, defined by
$\mu\ell_{c}^{-2}=e_{0}(\rho)$, i.e., $\ell_{c}\sim (\rho a)^{-1/2}$.
\end{itemize}
The length $\ell_{c}$ is sometimes called `correlation length' or
`healing length'. The name `uncertainty principle length'
is justified by the fact that this is the shortest length scale on
which the bosons can be localized without raising the energy per
particle above
$e_{0}$, according to the uncertainty principle. For dilute gases
$\rho a^3\ll 1$ and hence
\begin{equation}
a\ll \rho^{-1/3}\ll (\rho a^3)^{-1/6}\rho^{-1/3}\sim \ell_{c}.
\end{equation}
Bosons in their ground state are therefore `smeared out' over distances
large compared to the mean particle distance and their individuality
is entirely lost. Fermions, on the other hand, prefer to sit in
private rooms, i.e., $\ell_{c}$ can be comparable to $\rho^{-1/3}$.
In this respect the quantum nature of Bosons is much more pronounced
than for Fermions.
The three different length scales for Bosons will play a role in the
proof below.
Our lower bound for $e_{0}(\rho)$ is as follows.
\begin{theorem}[Lower bound in the thermodynamic limit]\label{lbth}
For a positive potential $v$ with finite range and $Y$ small enough
\begin{equation}\label{lowerbound}\frac{e_{0}(\rho)}{4\pi\mu\rho a}\geq
(1-C\,
Y^{1/17})
\end{equation}
with $C$ a constant. If $v$ does not have finite range, but decreases at
least as fast as
$1/r^{3+\varepsilon}$ at infinity with some $\varepsilon>0$, then an analogous
bound to (\ref{lowerbound})
holds, but with $C$ replaced by another constant and 1/17 by another
exponent, both of which may depend on $\varepsilon$.
\end{theorem}
It should be noted right away that the error term $-C\, Y^{1/17}$ in
(\ref{lowerbound}) is of no fundamental significance and is
not believed to reflect the true state of affairs. Presumably, it
does not even have the right sign. We mention in passing that
$C$ can be taken to be
$8.9$ \cite{S1999}.
As mentioned in the Introduction a lower bound on $E_{0}(N,L)$ for
finite $N$ and $L$ is of importance for applications to inhomogeneous
gases, and in fact we derive (\ref{lowerbound}) from such a bound. We
state it in the following way:
\begin{theorem}[Lower bound in a finite box] \label{lbthm2}
For a positive potential $v$ with finite range there is
a $\delta>0$ such that the the ground state energy of (\ref{ham}) with Neumann
conditions satisfies
\begin{equation}\label{lowerbound2}E_{0}(N,L)/N\geq 4\pi\mu\rho
a \left(1-C\,
Y^{1/17}\right)
\end{equation}
for all $N$ and $L$ with $YC'Y^{-6/17}$. Here
$C$ and $C'$ are constants,
independent of $N$ and $L$. (Note that the condition on $L/a$
requires in particular that $N$ must be large enough,
$N>\hbox{\rm (const.)}Y^{-1/17}$.)
As in Theorem \ref{lbth} such a bound, but possibly with other
constants and another
exponent for $Y$, holds also for potentials $v$ of infinite range
decreasing faster than $1/r^3$ at infinity.
\end{theorem}
The first step in the proof of (\ref{lbthm2}) is a generalization of
a lemma of Dyson, which allows us to replace $v$ by a `soft' potential,
at the cost of sacrificing kinetic energy and increasing the
effective range.
\begin{lemma}\label{dysonl} Let $v(r)\geq 0$ with finite range $R_{0}$. Let
$U(r)\geq 0$
be any function satisfying $\int U(r)r^2dr\leq 1$ and $U(r)=0$ for $rR_{0}$ the solution, $u_{0}$, satisfies $u_{0}(r)=r-a$
for $r>R_{0}$. By partial integration,
\begin{equation}\int_{0}^{R}\{\mu[u'_{0}(r)-(u_{0}(r)/r)]^2+
\mfr1/2v(r)|u_{0}(r)]^2\}dr=\mu a|R-a|^2/R^2.
\end
{equation}
But $|R-a|^2/R^2$ is precisely
the right side of (\ref{radial}) if $u$ satisfies the normalization condition.
This derivation of (\ref{dysonlemma}) for the special case (\ref{deltaU})
implies the
general case, because every $U$ can be written as a
superposition of $\delta$-functions,
$U(r)=\int R^{-2}\delta(r-R)\,U(R)R^2 dR$, and $\int U(R)R^2 dR\leq 1$
by assumption.
\hfill$\Box$
By dividing $\Lambda$ for given points $\x_{1},\dots,\x_{N}$ into Voronoi
cells ${\mathcal B}_{i}$ that contain all points
closer to $\x_{i}$ than to $\x_{j}$ with $j\neq i$ (these
cells are star shaped w.r.t. $\x_{i}$, indeed convex), the
following corollary of Lemma \ref{dysonl} can be derived in the same
way as the corresponding Eq.\ (28) in \cite{dyson}.
\begin{corollary} For any $U$ as in Lemma \ref{dysonl}
\begin{equation}\label{corollary}H_{N}\geq \mu a W\end{equation}
with
\begin{equation}\label{W}W(\x_{1},\dots,\x_{N})=\mfr1/2\sum_{i=1}^{N}U(t_{i}),
\end{equation}
where $t_{i}$ is the distance of $\x_{i}$ to its {\it nearest
neighbor} among the other points $\x_{j}$, $j=1,\dots, N$, i.e.,
\begin{equation}t_{i}(\x_{1},\dots,\x_{N})=\min_{j,\,j\neq
i}|\x_{i}-\x_{j}|.\end{equation}
\end{corollary}
\noindent
(Note that $t_{i}$ has here a slightly different meaning than in
(\ref{form}), where it denoted the distance to the nearest neighbor
among the $\x_{j}$ with $j\leq i-1$.)
Dyson considers in \cite{dyson} a one parameter family of $U$'s that
is essentially the same as the following choice, which is convenient for the
present purpose:
\begin{equation}U_{R}(r)=\begin{cases}\mfr1/3(R^3-R_{0}^3)^{-1}&\text{for
$R_{0}0$
\begin{equation}
H_{N}=\varepsilon H_{N}+(1-\varepsilon)H_{N}\geq \varepsilon
T_{N}+(1-\varepsilon)H_{N}
\end{equation}
with $T_{N}=-\sum_{i}\Delta_{i}$ and use (\ref{corollary}) only for the
part $(1-\varepsilon)H_{N}$. This gives
\begin{equation}\label{halfway}H_{N}\geq \varepsilon T_{N}+(1-\varepsilon)\mu
a
W_R.\end{equation}
We consider the operator on the right side
from the viewpoint of first order perturbation theory,
with $\varepsilon T_{N}$ as the unperturbed part, denoted $H_{0}$.
The ground state of $H_{0}$ in a box of side length $L$ is
$\Psi_{0}(\x_{1},\dots,\x_{N})\equiv L^{-3N/2}$ and we denote
expectation values in this state by $\langle\cdot\rangle_{0}$.
A computation, cf.\ Eq.\ (21) in \cite{LY1998}, gives
\begin{eqnarray}\label{firstorder}4\pi\rho\left(1-\mfr1/N\right)&\geq&
\langle W_R\rangle_{0}/N\\ &\geq& 4\pi\rho
\left(1-\mfr1/N\right)\left(1-\mfr{2R}/L\right)^3
\left(1+4\pi\rho(1-\mfr1/N)(R^3-R_{0}^3)/3)\right)^{-1}.\nonumber
\end{eqnarray}
The rationale behind the various factors is as follows: $(1-\mfr1/N)$ comes
from
the fact that the number of pairs is $N(N-1)/2$ and not $N^2/2$,
$(1-{2R}/L)^3$
takes into account the fact that the particles do not interact beyond the
boundary of
$\Lambda$, and the last factor measures the probability to find another
particle
within the interaction range of the potential $U_R$ for a given particle.
The first order result (\ref{firstorder}) looks at first sight quite
promising, for if we let
$L\to \infty$, $N\to \infty$ with $\rho=N/L^3$ fixed, and
subsequently take
$R\to\infty$, then $\langle W_R\rangle_{0}/N$ converges to $4\pi\rho$, which
is
just what is desired.
But the first order result (\ref{firstorder}) is not a
rigorous bound on $E_0(N,L)$, we need
{\it error estimates}, and these will depend on $\varepsilon$, $R$
and $L$.
We now recall {\it Temple's inequality} \cite{TE} for the expectations
values of an operator $H=H_{0}+V$ in the ground state
$\langle\cdot\rangle_{0}$ of $H_{0}$. It is a simple
consequence of the operator inequality
\begin{equation}(H-E_{0})(H-E_{1})\geq 0\end{equation}
for the two lowest eigenvalues, $E_{0}0$.
Furthermore, if $V\geq 0$ we may use $E_{1}\geq E_{1}^{(0)}$= second lowest
eigenvalue of $H_{0}$ and replace $E_{1}$ in (\ref{temple}) by $E_{1}^{(0)}$.
{}From (\ref{temple}) we get the estimate
\begin{equation}\label{estimate2}\frac{E_{0}(N,L)}{ N}\geq 4\pi \mu a\rho
\left(1-{\mathcal
E}(\rho,L,R,\varepsilon)\right)\end{equation}
with
\begin{equation}\label{error}1-{\mathcal
E}(\rho,L,R,\varepsilon)=(1-\varepsilon)\left(1-\frac1{\rho
L^3}\right)\left(1-\frac{\mu a\big(\langle
W_R^2\rangle_0-\langle W_R\rangle_0^2\big)}{\langle
W_R\rangle_0\big(E_{1}^{(0)}-\mu a\langle W_R\rangle_0\big)}\right).
\end{equation}
To evaluate this further one may use the estimates (\ref{firstorder}) and the
bound
\begin{equation}\label{square}
\langle W_R^2\rangle_0\leq 3\frac N{R^3-R_0^3}\langle W_R\rangle_0
\end{equation}
which follows from $U_R^2=3^{-1}({R^3-R_0^3})^{-1}U_R$ together with the
Cauchy-Schwarz
inequality. A glance at the form of the error term reveals, however, that it
is
{\it not} possible here to take the thermodynamic limit $L\to\infty$ with
$\rho$
fixed:
We have $E_{1}^{(0)}=\varepsilon\pi\mu/L^2$ (this is the kinetic energy of a
{\it single} particle in the first excited state in the box), and the factor
$E_{1}^{(0)}-\mu a\langle W_R\rangle_0$ in the denominator in (\ref{error})
is,
up to unimportant constants and lower order terms, $\sim (\varepsilon
L^{-2}-a\rho^2L^3)$. Hence the denominator eventually becomes negative and
Temple's inequality looses its validity if $L$ is large enough.
As a way out of this dilemma we divide the big box $\Lambda$ into cubic {\it
cells} of side length $\ell$ that is kept {\it fixed} as $L\to \infty$. The
number of cells, $L^3/\ell^3$, on the other hand, increases with $L$. The $N$
particles are distributed among these cells, and we use (\ref{error}), with
$L$
replaced by $\ell$, $N$ by the particle number, $n$, in a cell and $\rho$ by
$n/\ell^3$, to estimate the energy in each cell with {\it Neumann} conditions
on the boundary. This boundary condition leads to lower energy than any other
boundary condition. For each distribution of the particles we add the
contributions from the cells, neglecting interactions across boundaries.
Since
$v\geq 0$ by assumption, this can only lower the energy. Finally, we minimize
over all possible choices of the particle numbers for the various cells
adding up to $N$. The energy obtained in this way is a lower bound to
$E_0(N,L)$,
because we are effectively allowing discontinuous test functions for the
quadratic form given by $H_N$.
In mathematical terms, the cell method leads to
\begin{equation}\label{sum}
E_0(N,L)/N\geq(\rho\ell^3)^{-1}\inf \sum_{n\geq 0}c_nE_0(n,\ell)
\end{equation}
where the infimum is over all choices of coefficients $c_n\geq 0$ (relative
number of cells containing exactly $n$ particles), satisfying the constraints
\begin{equation}\label{constraints}
\sum_{n\geq 0}c_n=1,\qquad \sum_{n\geq 0}c_n n=\rho\ell^3.
\end{equation}
The minimization problem for the distributions of the particles among the
cells would be easy if we knew that the ground state energy $E_0(n,\ell)$ (or
a
good
lower bound to it) were convex in $n$. Then we could immediately conclude
that
it is best to have the particles as evenly distributed among the boxes as
possible, i.e., $c_n$ would be zero except for the $n$ equal to the
integer closest to
$\rho\ell^3$. This would give
\begin{equation}\label{estimate3}\frac{E_{0}(N,L)}{ N}\geq 4\pi \mu a\rho
\left(1-{\mathcal E}(\rho,\ell,R,\varepsilon)\right)\end{equation} i.e.,
replacement of $L$ in (\ref{estimate2}) by $\ell$, which is independent of
$L$.
The blow up of ${\mathcal E}$ for $L\to\infty$ would thus be avoided.
Since convexity of $E_0(n,\ell)$ is not known (except in the thermodynamic
limit)
we must resort to other means to show that $n=O(\rho\ell^3)$ in all
boxes. The rescue
comes from {\it superadditivity} of $E_{0}(n,\ell)$, i.e., the property
\begin{equation}\label{superadd}
%
E_0(n+n',\ell)\geq E_0(n,\ell)+E_0(n',\ell)
%
\end{equation}
which follows immediately from $v\geq 0$ by dropping the interactions between
the $n$ particles and the $n'$ particles. The bound (\ref{superadd}) implies
in
particular that for any $n,p\in{\mathbb N}$ with $n\geq p$
\begin{equation}\label{superadd1}
E(n,\ell)\geq [n/p]\,E(p,\ell)\geq \frac n{2p}E(p,\ell)
\end{equation}
since the largest integer $[n/p]$ smaller than $n/p$ is in any case $\geq
n/(2p)$.
The way (\ref{superadd1}) is used is as follows:
Replacing $L$ by $\ell$, $N$ by $n$ and $\rho$ by $n/\ell^3$ in
(\ref{estimate2}) we have for fixed $R$ and $\varepsilon$
\begin{equation}\label{estimate4}
E_{0}(n,\ell)\geq\frac{ 4\pi \mu a}{\ell^3}n(n-1)K(n,\ell)
\end{equation}
with a certain function $K(n,\ell)$ determined by (\ref{error}). We
shall see that $K$ is monotonously decreasing in $n$, so that if
$p\in{\mathbb N}$ and $n\leq p$ then
\begin{equation}\label{n